Question

A string is wrapped several times around the rim of a small hoop with radius 7cm and mass 2kg. The free end of the string is held in place and the hoop is released from rest.
After the hoop has descended 80cm, calculate (a) the angular speed of the hoop and (b) the speed of its center.

Answer 1

Answer:

Explanation:

Given that,

The radius of loop r = 7cm = 0.07m

Mass of hoop M = 2kg

The hoop is released from rest, the initial velocity is 0m/s

A. Angular speed of the hoop after a descended of h= 80cm = 0.8m

Applying the conservation of energy

Ei + W = Ef

Where,

Ei = initial energy of the system, I.e the initial potential and kinetic energy

Ef = final energy of the system I.e the final potential and kinetic energy

W = 0, since no external force is acting on the systems

Ui + K.Ei + 0 = Uf + K.Ef

System was initially at rest, K.Ei = 0

Uf = 0 at the zero level

Then,

Mgh = ½ •Icm•w² + ½M•Vcm²

Icm = Mr², since it is circular loop

Then,

Mgh = ½ •Mr²•w² + ½M•Vcm²

M cancels out

gh = ½ •r²•w² + ½ Vcm²

Since Vcm = wr

Then, gh = ½ •r²•w² + ½ w²r²

gh = r²w²

w = √(gh/r²)

w = √(9.81 × 0.8/0.07²)

w = 40 rad/s

b. Speed at the center

Since Vcm = wr

Vcm = 40×0.07

Vcm = 2.8 m/s